Answer:
C
Explanation:
The rate of a reaction decreases as time progresses. ... If the amount of reactant particles is decreasing as the reaction progresses, then the chance of successful collisions must also decrease, and ultimately when all the reactant particles have reacted, the reaction must stop and the rate become zero.
Answer:- 
Solution:- First of all we calculate the heat absorbed or released when the solute is added to the solvent. Here the solute is LiCl and the solvent is water.
To calculate the heat absorbed or released we use the formula:

q = heat absorbed or released
m = mass of solution
s = specific heat capacity
and
= change in temperature
mass of solution = mass of solute + mass of solvent
mass of solution = 5.00 g + 100.0 g = 105.0 g
(note:- density of pure water is 1 g per mL so the mass is same as its volume)
= 33.0 - 23.0 = 10.0 degree C
s = 
Let's plug in the values in the formula and calculate q.
q = 
q = 4389 J
To calculate the enthalpy of solution that is
we convert q to kJ and divide by the moles of solute.
moles of solute = 
= 0.118 moles
q =
= 4.389 kJ

= 
Since the heat is released which is also clear from the rise in temperature of the solution, the sign of enthapy of solution will be negative.
So, 
I dont think with this much amount of information we can solve this...unless its an reversible reaction in that case free energy =0
.412 g Mn304/ 1mol * 1 mol Mn304/228.82(MM of Mn3O4 molecule)* 3 mol Mn/1 mol Mn3O4 = .00540 mol Mn *54.94 g/mol = .296 g Mn